Question 1 |
The transfer function of a phase lead compensator is given by
D(s)=\frac{3\left ( s+\frac{1}{3T} \right )}{\left ( s+\frac{1}{T} \right )}
The frequency (in rad/sec), at which \angle D(j\omega ) is maximum, is
D(s)=\frac{3\left ( s+\frac{1}{3T} \right )}{\left ( s+\frac{1}{T} \right )}
The frequency (in rad/sec), at which \angle D(j\omega ) is maximum, is
\sqrt{\frac{3}{T^2}} | |
\sqrt{\frac{1}{3T^2}} | |
\sqrt{3T} | |
\sqrt{3T^3} |
Question 1 Explanation:
T(s)=\frac{1+3Ts}{1+Ts}
Frequency at which \angle T(j\omega ) is maximum,
(\omega _m)=\frac{1}{T\sqrt{\alpha }}
\alpha =\frac{1}{1/3}=3
\omega _m=\frac{1}{T\sqrt{3}}=\sqrt{\frac{1}{3T^2}}
Frequency at which \angle T(j\omega ) is maximum,
(\omega _m)=\frac{1}{T\sqrt{\alpha }}
\alpha =\frac{1}{1/3}=3
\omega _m=\frac{1}{T\sqrt{3}}=\sqrt{\frac{1}{3T^2}}
Question 2 |
The transfer function C(s) of a compensator is given below.
C(s)=\frac{(1+\frac{s}{0.1})(1+\frac{s}{100})}{(1+s)(1+\frac{s}{10})}
The frequency range in which the phase (lead) introduced by the compensator reaches the maximum is
C(s)=\frac{(1+\frac{s}{0.1})(1+\frac{s}{100})}{(1+s)(1+\frac{s}{10})}
The frequency range in which the phase (lead) introduced by the compensator reaches the maximum is
0.1 \lt \omega \lt 1 | |
1 \lt \omega \lt 10 | |
10 \lt \omega \lt 100 | |
\omega \gt 100 |
Question 2 Explanation:
Pole zero diagram of compensator transfer function is shown below.

Maximum phase lead is between 0.1 and 1.
0.1 \lt \omega \lt 1

Maximum phase lead is between 0.1 and 1.
0.1 \lt \omega \lt 1
Question 3 |
For the network shown in the figure below, the frequency (in rad/s) at which the maximum phase lag occurs is, ___________.


0.15 | |
0.32 | |
0.66 | |
0.92 |
Question 3 Explanation:
Assuming
\begin{aligned} R_1T&=9\Omega \\ R_2&=1\Omega\\ \text{ We can write,} \\ \frac{V_o(s)}{V_{in}(s)}&=\frac{R_2+\frac{1}{sC}}{R_1+R_2+\frac{1}{sC}} \\ &=\frac{1+R_2\cdot Cs}{1+(R_1+R_2)Cs} \\ &=\frac{1+R_2\cdot Cs}{1+\left (\frac{R_1+R_2}{R_2} \right )R_2 Cs}\\ \text{Let, } R_2C&=T \\ \frac{R_1+R_2}{R_2}&=\beta \\ \text{Hence, } \\ \frac{V_o(s)}{V_{in}(s)}&=\frac{1+Ts}{(1+\beta Ts)} \\ \text{Which represent }& \text{a lag compensator}\\ \because \text{ Here, } T&=R_2C=1.1=1 sec \\ \beta &=\frac{1+9}{1}=10\\ \text{Maximum phase }& \text{lag occurs at frequency}\\ \omega _n&=\frac{1}{T\sqrt{\beta }} \\ &=\frac{1}{1\sqrt{10}} =0.316 rad/sec. \end{aligned}
\begin{aligned} R_1T&=9\Omega \\ R_2&=1\Omega\\ \text{ We can write,} \\ \frac{V_o(s)}{V_{in}(s)}&=\frac{R_2+\frac{1}{sC}}{R_1+R_2+\frac{1}{sC}} \\ &=\frac{1+R_2\cdot Cs}{1+(R_1+R_2)Cs} \\ &=\frac{1+R_2\cdot Cs}{1+\left (\frac{R_1+R_2}{R_2} \right )R_2 Cs}\\ \text{Let, } R_2C&=T \\ \frac{R_1+R_2}{R_2}&=\beta \\ \text{Hence, } \\ \frac{V_o(s)}{V_{in}(s)}&=\frac{1+Ts}{(1+\beta Ts)} \\ \text{Which represent }& \text{a lag compensator}\\ \because \text{ Here, } T&=R_2C=1.1=1 sec \\ \beta &=\frac{1+9}{1}=10\\ \text{Maximum phase }& \text{lag occurs at frequency}\\ \omega _n&=\frac{1}{T\sqrt{\beta }} \\ &=\frac{1}{1\sqrt{10}} =0.316 rad/sec. \end{aligned}
Question 4 |
The transfer function of a compensator is given as
G_c(s)=\frac{s+a}{s+b}
The phase of the above lead compensator is maximum at
G_c(s)=\frac{s+a}{s+b}
The phase of the above lead compensator is maximum at
\sqrt{2} rad/s | |
\sqrt{3} rad/s | |
\sqrt{6} rad/s | |
1/ \sqrt{3} rad/s |
Question 4 Explanation:
\omega _m=\sqrt{\omega _{c1} \times \omega _{c2}}
\;\;=\sqrt{1 times 2}=\sqrt{2} rad/sec.
\;\;=\sqrt{1 times 2}=\sqrt{2} rad/sec.
Question 5 |
The transfer function of a compensator is given as
G_c(s)=\frac{s+a}{s+b}
G_c(s) is a lead compensator if
G_c(s)=\frac{s+a}{s+b}
G_c(s) is a lead compensator if
a = 1, b = 2 | |
a = 3, b = 2 | |
a =- 3, b =- 1 | |
a = 3, b = 1 |
Question 5 Explanation:
G_c(s)=\frac{s+a}{s+b}
G_c(j\omega )=\frac{j\omega +a}{j\omega +b}
\angle G_c(j\omega )=tan^{-1}\frac{\omega }{a}-tan^{-1}\frac{\omega }{b}
\;\;\;= tan^{-1}\left ( \frac{\frac{\omega }{a}-\frac{\omega }{b}}{1+\frac{\omega ^2 }{ab}} \right )
For G_c(s) to be a lead compensator
\angle G_c(j\omega ) \gt 0
\frac{\omega }{a} \gt \frac{\omega }{b}
\Rightarrow b \gt a
Option (A) Satisfies the above equation.
G_c(j\omega )=\frac{j\omega +a}{j\omega +b}
\angle G_c(j\omega )=tan^{-1}\frac{\omega }{a}-tan^{-1}\frac{\omega }{b}
\;\;\;= tan^{-1}\left ( \frac{\frac{\omega }{a}-\frac{\omega }{b}}{1+\frac{\omega ^2 }{ab}} \right )
For G_c(s) to be a lead compensator
\angle G_c(j\omega ) \gt 0
\frac{\omega }{a} \gt \frac{\omega }{b}
\Rightarrow b \gt a
Option (A) Satisfies the above equation.
There are 5 questions to complete.